Law of Sines / Sine Rule

For any triangle ABC, as shown,

the law of sines or sine rule states that

We can also write the law of sines or sine rule as:

We can use the Law of Sines when solving triangles. Solving a triangle means to find the unknown lengths and angles of the triangle.

Case 1: Given Two Angles And One Side

We will first consider the situation when we are given 2 angles and one side of a triangle

Example:

Solve triangle PQR in which ∠ P = 63.5° and ∠ Q = 51.2° and r = 6.3 cm.

First, calculate the third angle.

∠ R = 180° – 63.5° – 51.2° = 65.3°

Next, calculate the sides.

∠ R = 65.3°, p = 6.21 cm and q = 5.40 cm

Case 2: Given Two Sides And One Angle

We will now consider the situation when we are given two sides and one angle of a triangle.

Ambiguous Case

If you are given two sides and a non-included acute angle and the side facing the given angle is less than the other side, you would obtain two sets of answers. The solution is said to be ambiguous.

Example:

Solve triangle PQR in which ∠ P = 56°, p = 10 cm and q = 12 cm

Solution:

Notice that we can construct two different triangles from the given information.

When ∠ Q = 95.8˚, ∠ R = 180˚ – 56˚ – 95.8˚ = 28.2˚

The two sets of solutions are:

∠ Q = 84.2°, ∠ R = 39.8°, r = 7.72 cm

∠ Q = 95.8°, ∠ R = 28.2°, r = 5.70 cm

Case 3: Given Two Sides And An Obtuse Angle

We will now consider the situation when we are given two sides and an obtuse angle of a triangle.

Example:

Solve ∆ PQR in which ∠ P =116°, p = 8.3 cm and q = 5.4 cm.

Solution:

Q cannot be an obtuse angle because the sum of angles in the triangle will exceed 180˚. The only valid value for Q is 35.8˚.

∠ Q = 35.8°, ∠ R = 180° – 116° – 35.8° = 28.2°

The solution is ∠ Q = 35.8° , ∠ R = 28.2° and r = 4.36 cm

Videos

The following video shows more examples of solving triangles using Law of Sines

Solving a triangle given two sides and one angle -
Professor Edward Burger explains solving a triangle given two sides and one angle

Solving a triangle (sas) - ambiguous case
Professor Edward Burger explains another example of solving an SAS triangle

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